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oksian1 [2.3K]
3 years ago
6

Find the value of -24/35÷-8/15​

Mathematics
2 answers:
kirill115 [55]3 years ago
6 0

Answer:

9/7

Step-by-step explanation:

-24/35 ÷ -8/15

=-24/35 × -15/8

=9/7

denpristay [2]3 years ago
5 0

Answer:

9/7

Step-by-step explanation:

when dividing with a fraction, the -8/15 reciprocates to 15/-8. Then, -8 cancels out -24 remaining with 3 since they are both negative while 15 and 35 have a common of 5 giving them a remainder of 3 and 7 respectively. you then multiply what you have which is 3/7 * 3/1 giving you 9/7.

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Solve the whole problem i give brainliest please help me.
Anna11 [10]

Answer:

The inequality is 12.5v + 70 ≥ 215

and the amount of visits they can make is 12 visits

Step-by-step explanation:

if you take away (subtract) 70 from both sides, you'll get

12.5v ≥ 145

and when you divide both sides by 12.5, you'll get 11.6, or 12

3 0
3 years ago
By ordering the data find the Lower Quartile<br> 5, 7, 4, 3, 2, 6, 8, 9, 10, 15 *
Klio2033 [76]

Answer:

.Lower quartile = 3.5

Step-by-step explanation:

Rearranging the data from lowest to highest.

2, 3, 4, 5, 6, 7, 8, 9 , 10, 15

First finding the median to make it easier to find the lower quartile.

median : 6 and 7 but right now they're irrelevant so we don't honestly need to work that out.

The lower quartile would be 3 and 4

3+4= 7 divide by 2 = 3.5 is your lower quartile

8 0
3 years ago
Not sure if any of this is correct, but it’s what I got so far
Irina18 [472]

Problem 1 is correct. You use the pythagorean theorem to find the hypotenuse.

==================================================

Problem 2 has the correct answer, but one part of the steps is a bit strange. I agree with the 132 ft/sec portion; however, I'm not sure why you wrote \frac{1 \text{ sec}}{132 \text{ ft}}=\frac{0.59\overline{09}}{78 \text{ ft}}*127 \text{ ft}

I would write it as \frac{1\text{ sec}}{132 \text{ ft}}*127 \text{ ft} = \frac{127}{132} \text{ sec} \approx 0.96 \text{ sec}

==================================================

For problem 3, we first need to convert the runner's speed from mph to feet per second.

17.5 \text{ mph} = \frac{17.5 \text{ mi}}{1 \text{ hr}}*\frac{1 \text{ hr}}{60 \text{ min}}*\frac{1 \text{ min}}{60 \text{ sec}}*\frac{5280 \text{ ft}}{1 \text{ mi}} \approx 25.667 \text{ ft per sec}

Since the runner needs to travel 90-12 = 78 ft, this means\text{time} = \frac{\text{distance}}{\text{speed}} \approx \frac{78 \text{ ft}}{25.667 \text{ ft per sec}} \approx 3.039 \text{ sec}

So the runner needs about 3.039 seconds. In problem 2, you calculated that it takes about 0.96 seconds for the ball to go from home to second base. The runner will not beat the throw. The ball gets where it needs to go well before the runner arrives there too.

-------------

The question is now: how much of a lead does the runner need in order to beat the throw?

Well the runner needs to get to second base in under 0.96 seconds.

Let's calculate the distance based on that, and based on the speed we calculated earlier above.

\text{distance} = \text{rate}*\text{time} \approx (25.667 \text{ ft per sec})*(0.96 \text{ sec}) \approx 24.64032 \text{ ft}

This is the distance the runner can travel if the runner only has 0.96 seconds. So the lead needed is 90-24.64032 = 65.35968 feet

This is probably not reasonable considering it's well over halfway (because 65.35968/90 = 0.726 = 72.6%). If the runner is leading over halfway, then the runner is probably already in the running motion and not being stationary.

As you can see, the runner is very unlikely to steal second base. Though of course such events do happen in real life. What may explain this is the reaction time of the catcher may add on just enough time for the runner to steal second base. For this problem however, we aren't considering the reaction time. Also, not all catchers can throw the ball at 90 mph which is quite fast. According to quick research, the MLB says the average catcher speed is about 81.8 mph. This slower throwing speed may account for why stealing second base isn't literally impossible, although it's still fairly difficult.

5 0
3 years ago
Mason plays a game by flipping two fair coins. He wins the game if both coins land facing heads up. If Mason plays 200 times, ho
aev [14]

Answer:

Total no.of possible outcomes = 200 × 4 = 800

No.of favourable outcomes = 200 × 1 = 200

P (E) = no.of favourable outcomes / total no.of possible outcomes

= 200/800

= 1/4

7 0
3 years ago
Read 2 more answers
Suppose you produce a camera that costs $50 to make. At a selling price of $45, you will _____.
Daniel [21]
You will lose money. Because $50 to make subtract $45 to sell (50-45) leaves you losing $5.
4 0
3 years ago
Read 2 more answers
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